Difference between <? super T> and <? extends T> in Java [duplicate]

Answer 1

You can go through all the answers above to understand why the .add() is restricted to '<?>', '<? extends>', and partly to '<? super>'.

But here's the conclusion of it all if you want to remember it, and dont want to go exploring the answer every time:

List<? extends A> means this will accept any List of A and subclass of A. But you cannot add anything to this list. Not even objects of type A.

List<? super A> means this will accept any list of A and superclass of A. You can add objects of type A and its subclasses.

Answer 2

Imagine having this hierarchy

1. Extends

By writing

    List<? extends C2> list;

you are saying that list will be able to reference an object of type (for example) ArrayList whose generic type is one of the 7 subtypes of C2 (C2 included):

1. C2: new ArrayList<C2>();, (an object that can store C2 or subtypes) or
2. D1: new ArrayList<D1>();, (an object that can store D1 or subtypes) or
3. D2: new ArrayList<D2>();, (an object that can store D2 or subtypes) or...

and so on. Seven different cases:

    1) new ArrayList<C2>(): can store C2 D1 D2 E1 E2 E3 E4
    2) new ArrayList<D1>(): can store    D1    E1 E2  
    3) new ArrayList<D2>(): can store       D2       E3 E4
    4) new ArrayList<E1>(): can store          E1             
    5) new ArrayList<E2>(): can store             E2             
    6) new ArrayList<E3>(): can store                E3             
    7) new ArrayList<E4>(): can store                   E4             

We have a set of "storable" types for each possible case: 7 (red) sets here graphically represented

As you can see, there is not a safe type that is common to every case:

• you cannot list.add(new C2(){}); because it could be list = new ArrayList<D1>();
• you cannot list.add(new D1(){}); because it could be list = new ArrayList<D2>();

and so on.

2. Super

By writing

    List<? super C2> list;

you are saying that list will be able to reference an object of type (for example) ArrayList whose generic type is one of the 7 supertypes of C2 (C2 included):

• A1: new ArrayList<A1>();, (an object that can store A1 or subtypes) or
• A2: new ArrayList<A2>();, (an object that can store A2 or subtypes) or
• A3: new ArrayList<A3>();, (an object that can store A3 or subtypes) or...

and so on. Seven different cases:

    1) new ArrayList<A1>(): can store A1          B1 B2       C1 C2    D1 D2 E1 E2 E3 E4
    2) new ArrayList<A2>(): can store    A2          B2       C1 C2    D1 D2 E1 E2 E3 E4
    3) new ArrayList<A3>(): can store       A3          B3       C2 C3 D1 D2 E1 E2 E3 E4
    4) new ArrayList<A4>(): can store          A4       B3 B4    C2 C3 D1 D2 E1 E2 E3 E4
    5) new ArrayList<B2>(): can store                B2       C1 C2    D1 D2 E1 E2 E3 E4
    6) new ArrayList<B3>(): can store                   B3       C2 C3 D1 D2 E1 E2 E3 E4
    7) new ArrayList<C2>(): can store                            C2    D1 D2 E1 E2 E3 E4

We have a set of "storable" types for each possible case: 7 (red) sets here graphically represented

As you can see, here we have seven safe types that are common to every case: C2, D1, D2, E1, E2, E3, E4.

• you can list.add(new C2(){}); because, regardless of the kind of List we're referencing, C2 is allowed
• you can list.add(new D1(){}); because, regardless of the kind of List we're referencing, D1 is allowed

and so on. You probably noticed that these types correspond to the hierarchy starting from type C2.

Notes

Here the complete hierarchy if you wish to make some tests

interface A1{}
interface A2{}
interface A3{}
interface A4{}

interface B1 extends A1{}
interface B2 extends A1,A2{}
interface B3 extends A3,A4{}
interface B4 extends A4{}

interface C1 extends B2{}
interface C2 extends B2,B3{}
interface C3 extends B3{}

interface D1 extends C1,C2{}
interface D2 extends C2{}

interface E1 extends D1{}
interface E2 extends D1{}
interface E3 extends D2{}
interface E4 extends D2{}

Answer 3

Adding an item to the list:

• List< ? extends X > doesn't allow to add anything, except for null into the list.

• List< ? super X > allows to add anything that is-a X (X or its subtype), or null.

Getting an item from the list:

• When you get an item from List< ? extends X >, you can assign it to a variable of type X or any supertype of X, including Object.
• When you get an item from List< ? super X >, you can only assign it to a variable of type Object.

Some examples:

    List<? extends Number> list1 = new ArrayList<Integer>();
    list1.add(null);  //OK
    Number n = list1.get(0);  //OK
    Serializable s = list1.get(0);  //OK
    Object o = list1.get(0);  //OK

    list1.add(2.3);  //ERROR
    list1.add(5);  //ERROR
    list1.add(new Object());  //ERROR
    Integer i = list1.get(0);  //ERROR

---

    List<? super Number> list2 = new ArrayList<Number>();
    list2.add(null);  //OK
    list2.add(2.3);  //OK
    list2.add(5);  //OK
    Object o = list2.get(0);  //OK

    list2.add(new Object());  //ERROR
    Number n = list2.get(0);  //ERROR
    Serializable s = list2.get(0);  //ERROR
    Integer i = list2.get(0);  //ERROR

Answer 4

Example, Order of inheritance is assumed as O > S > T > U > V

Using extends Keyword ,

Correct:

List<? extends T> Object = new List<T>();
List<? extends T> Object = new List<U>();
List<? extends T> Object = new List<V>();

InCorrect:

List<? extends T> Object = new List<S>();
List<? extends T> Object = new List<O>();

super Keyword:

Correct:

List<? super T> Object = new List<T>();
List<? super T> Object = new List<S>();
List<? super T> Object = new List<O>();

InCorrect:

List<? super T> Object = new List<U>();
List<? super T> Object = new List<V>();

Adding object: List Object = new List();

Object.add(new T()); //error

But Why error ? Let's look at the Possibilities of initializations of List Object

List<? extends T> Object = new List<T>();
List<? extends T> Object = new List<U>();
List<? extends T> Object = new List<V>();

If we use Object.add(new T()); then it will be correct only if

List<? extends T> Object = new List<T>(); 

But there are extra two possibilities

List Object = new List(); List Object = new List(); If we try to add (new T()) to the above two possibilities it will give an error because T is the superior class of U and V . we try to add a T object [which is (new T()) ] to List of type U and V . Higher class object(Base class) cannot be passed to lower class Object(Sub class).

Due to the extra two possibilities , Java gives you error even if you use the correct possilibity as Java don't know what Object you are referring to .So you can't add objects to List Object = new List(); as there are possibilities that are not valid.

Adding object: List Object = new List();

Object.add(new T()); // compiles fine without error
Object.add(new U()); // compiles fine without error
Object.add(new V()); // compiles fine without error

Object.add(new S()); //  error
Object.add(new O()); //  error

But why error occurs in the above two ? we can use Object.add(new T()); only on the below possibilities,

List<? super T> Object = new List<T>();
List<? super T> Object = new List<S>();
List<? super T> Object = new List<O>();

If we Tried to use Object.add(new T()) in List Object = new List(); and List Object = new List(); then it will give error This is because We can't add T object[which is new T()] to the List Object = new List(); because it is an object of type U . We can't add a T object[which is new T()] to U Object because T is a base class and U is a sub class . We can't add base class to subclass and that's why error occurs . This is same for the another case .

Answer 5

The most confusing thing here is that whatever type restrictions we specify, assignment works only one way:

baseClassInstance = derivedClassInstance;

You may think that Integer extends Number and that an Integer would do as a <? extends Number>, but the compiler will tell you that <? extends Number> cannot be converted to Integer (that is, in human parlance, it is wrong that anything that extends number can be converted to Integer):

class Holder<T> {
    T v;
    T get() { return v; }
    void set(T n) { v=n; }
}
class A {
    public static void main(String[]args) {
        Holder<? extends Number> he = new Holder();
        Holder<? super Number> hs = new Holder();

        Integer i;
        Number n;
        Object o;

        // Producer Super: always gives an error except
        //       when consumer expects just Object
        i = hs.get(); // <? super Number> cannot be converted to Integer
        n = hs.get(); // <? super Number> cannot be converted to Number
                      // <? super Number> cannot be converted to ... (but
                      //       there is no class between Number and Object)
        o = hs.get();

        // Consumer Super
        hs.set(i);
        hs.set(n);
        hs.set(o); // Object cannot be converted to <? super Number>

        // Producer Extends
        i = he.get(); // <? extends Number> cannot be converted to Integer
        n = he.get();
        o = he.get();

        // Consumer Extends: always gives an error
        he.set(i); // Integer cannot be converted to <? extends Number>
        he.set(n); // Number cannot be converted to <? extends Number>
        he.set(o); // Object cannot be converted to <? extends Number>
    }
}

hs.set(i); is ok because Integer can be converted to any superclass of Number (and not because Integer is a superclass of Number, which is not true).

EDIT added a comment about Consumer Extends and Producer Super -- they are not meaningful because they specify, correspondingly, nothing and just Object. You are advised to remember PECS because CEPS is never useful.

Answer 6

When to use extends and super

Wildcards are most useful in method parameters. They allow for the necessary flexibility in method interfaces.

People are often confused when to use extends and when to use super bounds. The rule of thumb is the get-put principle. If you get something from a parametrized container, use extends.

int totalFuel(List<? extends Vehicle> list) {
int total = 0;
for(Vehicle v : list) {
    total += v.getFuel();
}
return total;}

The method totalFuel gets Vehicles from the list, asks them about how much fuel they have, and computes the total. If you put objects into a parameterized container, use super.

int totalValue(Valuer<? super Vehicle> valuer) {
int total = 0;
for(Vehicle v : vehicles) {
    total += valuer.evaluate(v);
}
return total;}

The method totalValue puts Vehicles into the Valuer. It's useful to know that extends bound is much more common than super.